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I have a problem with understanding how the resolution of the identity of an operator is presented in some literature for physicists.

I'm a student of mathematics, and I understand the notion of a spectral measure (which is somethimes called the resolution of identity) and also have some knowledge in spectral theory (for normal operators).

Here is my brief explanation what do I understand:

Let $H$ be Hilbert space (with an inner product linear w.r.t 2nd coordinate) with an orthonormal basis $(e_{n})$ and define a linear operator (diagonal operator) $$ A = \sum_{i \geq 1} \lambda_i \left|e_i\right>\left<e_i\right|,$$ where $(\lambda_i)$ is a sequence of complex numbers (the properties of this sequences determine the properties of $A$ such as boundedness, selfadjointness, compactness etc.), to make sense of $A$ we assume that the above series converges in SOT, I also used Dirac bra-ket notation. The associated spectral measure of $A$ is defined via $$ E(\Delta) = \sum_i \mathbf{1}_{\Delta}(\lambda_i) \left|e_i\right>\left<e_i\right|,$$ where $\Delta$ is an element of Borel sigma field over the spectrum of $A$, and we have that $$\left<x| Ay\right> = \int_{\sigma(A)} \lambda \left<x| E(d\lambda)y\right> \ \ (x,y \in H).$$

Very often physicists would use the following notation for $A$ which acts on an element $\psi \in H$ $$ A\left|\psi\right> = \sum_{i} \lambda_i \left|i\right>\left<i\right|\left|\psi\right>.$$

My problems with notation

I started reading some notes, books about quantum field theory, and often it is written that, the identity operator $I$ on some (separable) Hilbert space $H$, has the expansion, called the resolution of the identity $$ I= \int dq^{\prime} \left|q^{\prime}\right>\left<q^{\prime}\right|,$$ I don't know whether it matters here but $\{\left|q\right>\}$ is supposed to be a complete set of states.

Reference: http://eduardo.physics.illinois.edu/phys582/582-chapter5.pdf bottom of p. 129.

My question

Is the notion of the above resolution of the identity the same as an integral w.r.t a spectral measure ($I$ is a diagonal operator)? If yes, how should I understand the above notation. If no, what do they actually mean by this resolution of the identity and how do they define this integral. I noticed that in a lot of book concerning quantum mechanics there are many calculations, but not very many definitions and assumptions, which makes stuff hard to understand for a mathematician.

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The answer is Yes. The interpretation of the notation is quite straight forward: $dq'|q'\rangle\langle q'| = E(dq')$. We need to presume that $E$ is the spectral measure of an operator $Q' = \int q' E(dq') = \int dq'\, q' |q'\rangle\langle q'|$. The only aspect that doesn't necessarily mesh well with your question, as written, is the fact that you've defined a spectral measure $E$ only for operators whose spectrum consists of eigenvalues, since $A|e_i\rangle = \lambda_i |e_i\rangle$. Spectral measures can be defined for operators with any kind of spectrum, including continuous.

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  • $\begingroup$ Thanks, I know that you have spectral theorem for any normal operator defined on a Hilbert space, however the notation for the resolution of identity in the example which I gave about was for the identity operator, which is in fact diagonal. One quick question so it supposed to be $$I= \int d q^{\prime} q^{\prime} \left|q^{\prime} \right> \left<q^{\prime}\right|$$ instead of $$I= \int d q^{\prime} \left|q^{\prime} \right> \left<q^{\prime}\right|$$? $\endgroup$ – Eric Feb 6 '15 at 16:06
  • $\begingroup$ Your first formula gives $Q'$ and not $I$, with $I$ given correctly by your second formula. The prototypical example of $Q'$ is the position operator in quantum mechanics. It has a simple continuous spectrum (in 1-dimension, that is) and that is why its "eigenvectors" are convenient for writing down a resolution of the identity. Essentially, you are representing identity using functional calculus $I = f(Q) = \int dq' f(q') |q'\rangle\langle q'|$, where $f(x) \equiv 1$. $\endgroup$ – Igor Khavkine Feb 6 '15 at 19:32
  • $\begingroup$ A good rule of thumb in interpreting this notation is to make sure that your interpretation works when the Hilbert space is finite dimensional and all integrals are replaced by a sums. $\endgroup$ – Igor Khavkine Feb 6 '15 at 19:34
  • $\begingroup$ Great answer! Thank you. For the identity we even don't need the functional calculus, we know that $Q^{\prime}$ as a selfadjoint operator admits a unique spectral measure $E$, thus by using the properties of a spectral measure we got $I=E(\sigma(Q^{\prime}))= \int_{\sigma(Q^{\prime})} 1 E(d\lambda)$, which can be of course written in a different notation which involves bra and kets. $\endgroup$ – Eric Feb 6 '15 at 19:47
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See $\S$ 4.4 of de Madrid's "The role of the rigged Hilbert space in quantum mechanics"

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